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Question

Solve $$\log _{0.3}\left ( x^{2}-x+1 \right )> 0$$.


Solution

$$\log _{0.3}\left ( x^{2}-x+1 \right )> 0$$ or $$0< x^{2}-x+1< \left ( 0.3 \right )^{0}$$
or   $$0< x^{2}-x+1< 1\Rightarrow x^{2}-x+1> 0$$ and $$x^{2}-x< 0$$
or   $$x\left ( x-1 \right )< 0$$
$$\Rightarrow $$   $$0< x< 1$$  .... (as $$x^{2}-x+1=\left ( x-\dfrac{1}{2} \right )^{2}+\dfrac{3}{4}> 0$$), for all real x
Ans: 1

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